Let $f$ be a real valued continuous algebraic function on $\mathbb R^n$. Suppose the zero set of $f$ is bounded, i.e., if $x$ is large enough, $f(x)\neq 0$. Is there any estimate of the sort $f(x)\ge c(1+x)^{N}$ for some $c>0$ and $n\in\mathbb N$? A prototypical example of such an inequality is $\sqrt{1+x^2}x\ge c(1+x)^{1}$, but I would appreciate general estimates along this line that apply to algebraic functions of $n$ variables.

$\begingroup$ Maybe you can have radial bounds on halflines $x=\lambda x_0, x>>1$, $x_0$ fixed, by looking at the corresponding Puiseux series in the variable $\lambda$ near $\infty$. $\endgroup$ – Loïc Teyssier Dec 15 '15 at 18:31

$\begingroup$ @LoïcTeyssier If I want to do it In several variables, I want to make the estimate uniform, but I am unable to see why the Puiseux expansion is continuous w.r.t. $x_0$. $\endgroup$ – Fan Zheng Dec 15 '15 at 18:34

$\begingroup$ I am well aware of that, I was only suggesting this in case it'd be sufficient for your purpose to have the estimate on lines. Yet, in my experience, there is a kind of «semicontinuity» arising in that problems. $\endgroup$ – Loïc Teyssier Dec 15 '15 at 18:37

1$\begingroup$ No, if there are no continuity assumptions. Otherwise, e.g., you may define $f(x,y)$ to be the root of $f^2+4(x^2+y^2)f+(x^2y^2)=0$ of minimal absolute value if it is nonzero, and the other root otherwise. $\endgroup$ – Ilya Bogdanov Dec 22 '15 at 12:39
Your function $f$ is, in particular, a continuous semialgebraic function on $\mathbf R^n$, i.e., its graph is a semialgebraic subset of $\mathbf R^{n+1}$. Such functions are known to have subpolynomial growth. More precisely (cf. Bochnak, Coste, Roy: Real algebraic geometry, Proposition 2.6.2, p. 43):
Let $S\subseteq\mathbf R^n$ be a closed semialgebraic subset, and let $f\colon S\rightarrow \mathbf R$ be a continuous semialgebraic function. Then there are $c\in\mathbf R$ and $N\in\mathbf N$ such that $$ f(x)\leq c(1+x^2)^N $$ for all $x\in S$.
Apply this to the inverse $1/f$ of your function on the closed semialgebraic subset $x\geq A$, where $A\in\mathbf R$ is such that all zeros of $f$ have norm $<A$.

$\begingroup$ A belated thank you: found the (very informative) reference. $\endgroup$ – Fan Zheng Oct 16 '16 at 23:04
Update: As pointed out by Igor, this is not convincing. I only show that if $p(x,f)=0$ and $p(x,0)\not=0$ for $x>R$, then the claim holds. However, the OP only assumes that $f\not=0$ for $x>R$ (I mixed up the two originally).
Original answer: Yes, this works. If we had $f(x_N)\lesssim x_N^{N}$ for arbitrarily large $N$, then the constant coefficient $p=a_0$ in the equation $\sum a_jf^j=0$ solved by $f$ would have the same property.
So it suffices to show that if a polynomial satisfies $p(x)>0$ for $x>R$, then in fact $p(x)\gtrsim x^{N}$ for some $N$. This follows from the Positivstellensatz, which says, in our setting, that $$ p(x) = \frac{1 + (x^2R^2)F(x)+G(x)}{(x^2R^2)H(x)+K(x)} \ge \frac{1}{(x^2R^2)H(x)+K(x)} \gtrsim x^{N} ; $$ here, $F,G, H, K$ are sums of squares of polynomials.
(I suspect there must be a much more elementary argument.)


$\begingroup$ Ah, I see. I was thinking that if $y$ is a solution of $p(x, y) = 0,$ then it is also a solution of $y p(x, y)=0,$ without adding the $y=0$ component. $\endgroup$ – Igor Rivin Dec 15 '15 at 21:21
Well it's true in dimension $1$, and I think the following argument should reduce the case $n>1$ to the $n=1$ statement.
Let $r_0 = \max_{f(x) = 0} \ f(x) \$. Then $f$ has constant sign on $\ x \ > r_0$ (continuous function on a connected set), and we may assume the sign is positive (else replace $f$ by $f$). For $r \geq r_0$, define $$ F(r) := \min_{\x\ = r} f(x). $$ This is an algebraic function of $r$ because it is the value of the algebraic function $f(x_1,\ldots,x_n)$ on the algebraic variety (typically a curve) in ${\bf R}^{n+2}$ specified by the following equations on $r$, $x_1,\ldots,x_n$, and $c$: $$ \sum_{i=1}^n x_i^2 = r^2, \quad \forall i: \frac{\partial f}{\partial x_i} = c x_i. $$ But by hypothesis $F(r) \neq 0$ for $r > r_0$, so there exists $N<\infty$ such that $f(r) \gg r^{N}$ for large r, as claimed.